Question

When equal weights of methyl alcohol and ethyl alcohol react with excess of sodium metal,the volume of H2 liberated is more in the case of
1) C2H5OH 2) CH3OH 3) Equal in both 4) H2 is not liberated

Answer 1

Answer:

If a small piece of sodium is dropped into ethanol, it reacts steadily to give off bubbles of hydrogen gas and leaves a colorless solution of sodium ethoxide:  CH3CH2ONa . The anion component is an alkoxide.

2CH3CH2OH(l)+2Na(s)→2CH3CH2O−(aq)+2Na+(aq)+H2(g)(1)

If the solution is evaporated carefully to dryness, then sodium ethoxide ( CH3CH2ONa ) is left behind as a white solid. Although initially this appears as something new and complicated, in fact, it is exactly the same (apart from being a more gentle reaction) as the reaction between sodium and water - something you have probably known about for years.

2H2O(l)+2Na(s)→2OH−(aq)+2Na+(aq)+H2(g)(2)

If the solution is evaporated carefully to dryness, then the sodium hydroxide ( NaOH ) is left behind as a white solid.

We normally, of course, write the sodium hydroxide formed as  NaOH  rather than  HONa  - but that's the only difference. Sodium ethoxide is just like sodium hydroxide, except that the hydrogen has been replaced by an ethyl group. Sodium hydroxide contains  OH−  ions; sodium ethoxide contains  CH3CH2O−  ions.

Explanation: